Integrand size = 12, antiderivative size = 77 \[ \int \frac {1}{(c \sin (a+b x))^{5/2}} \, dx=-\frac {2 \cos (a+b x)}{3 b c (c \sin (a+b x))^{3/2}}+\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right ),2\right ) \sqrt {\sin (a+b x)}}{3 b c^2 \sqrt {c \sin (a+b x)}} \]
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Time = 0.02 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2716, 2721, 2720} \[ \int \frac {1}{(c \sin (a+b x))^{5/2}} \, dx=\frac {2 \sqrt {\sin (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right ),2\right )}{3 b c^2 \sqrt {c \sin (a+b x)}}-\frac {2 \cos (a+b x)}{3 b c (c \sin (a+b x))^{3/2}} \]
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Rule 2716
Rule 2720
Rule 2721
Rubi steps \begin{align*} \text {integral}& = -\frac {2 \cos (a+b x)}{3 b c (c \sin (a+b x))^{3/2}}+\frac {\int \frac {1}{\sqrt {c \sin (a+b x)}} \, dx}{3 c^2} \\ & = -\frac {2 \cos (a+b x)}{3 b c (c \sin (a+b x))^{3/2}}+\frac {\sqrt {\sin (a+b x)} \int \frac {1}{\sqrt {\sin (a+b x)}} \, dx}{3 c^2 \sqrt {c \sin (a+b x)}} \\ & = -\frac {2 \cos (a+b x)}{3 b c (c \sin (a+b x))^{3/2}}+\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right ),2\right ) \sqrt {\sin (a+b x)}}{3 b c^2 \sqrt {c \sin (a+b x)}} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(c \sin (a+b x))^{5/2}} \, dx=-\frac {2 \left (\cos (a+b x)+\operatorname {EllipticF}\left (\frac {1}{4} (-2 a+\pi -2 b x),2\right ) \sin ^{\frac {3}{2}}(a+b x)\right )}{3 b c (c \sin (a+b x))^{3/2}} \]
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Time = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.36
method | result | size |
default | \(-\frac {\sqrt {-\sin \left (b x +a \right )+1}\, \sqrt {2 \sin \left (b x +a \right )+2}\, \left (\sin ^{\frac {5}{2}}\left (b x +a \right )\right ) F\left (\sqrt {-\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{3}\left (b x +a \right )\right )+2 \sin \left (b x +a \right )}{3 c^{2} \sin \left (b x +a \right )^{2} \cos \left (b x +a \right ) \sqrt {c \sin \left (b x +a \right )}\, b}\) | \(105\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.65 \[ \int \frac {1}{(c \sin (a+b x))^{5/2}} \, dx=\frac {{\left (\sqrt {2} \cos \left (b x + a\right )^{2} - \sqrt {2}\right )} \sqrt {-i \, c} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + {\left (\sqrt {2} \cos \left (b x + a\right )^{2} - \sqrt {2}\right )} \sqrt {i \, c} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + 2 \, \sqrt {c \sin \left (b x + a\right )} \cos \left (b x + a\right )}{3 \, {\left (b c^{3} \cos \left (b x + a\right )^{2} - b c^{3}\right )}} \]
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\[ \int \frac {1}{(c \sin (a+b x))^{5/2}} \, dx=\int \frac {1}{\left (c \sin {\left (a + b x \right )}\right )^{\frac {5}{2}}}\, dx \]
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\[ \int \frac {1}{(c \sin (a+b x))^{5/2}} \, dx=\int { \frac {1}{\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]
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\[ \int \frac {1}{(c \sin (a+b x))^{5/2}} \, dx=\int { \frac {1}{\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {1}{(c \sin (a+b x))^{5/2}} \, dx=\int \frac {1}{{\left (c\,\sin \left (a+b\,x\right )\right )}^{5/2}} \,d x \]
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